Derivative of $\frac{d}{dt}\ln(6t^2+9t+12)=$

Question

$\dfrac{d}{dt}\ln(6t^2+9t+12)=$

$y=2\ln(6t)+\ln(9t)+\ln(12)$

$y\;'=2\dfrac{1}{6t}(6)+\dfrac{1}{9t}(9)+0$

$=\dfrac{12}{6t}+\dfrac{9}{9t}=\dfrac{2}{t}+\dfrac{1}{t}$

What am I doing wrong?

Answer 1

You cannot simplify a logarithm like you did in the second step - that is, $$y=\ln{(6t^2+9t+12)}$$ does not equal $$y=2\ln{6t^2}+\ln{9t}+\ln{12}$$ You can split up logarithms when you are multiplying numbers - that is, $$\ln(x\times x)=\ln(x)+\ln(x)$$ In this case, though, just use the chain rule: $$f'(g(x))=f'(g(x))g'(x)$$

Because $g(t)=6t^2+9t+12$, $$f'(t)=y'=(12t+9)\left(\frac{1}{6t^2+9t+12}\right)$$ which becomes $$f'(t)=\frac{12t+9}{6t^2+9t+12}$$ and then $$f'(t)=\frac{4t+3}{2t^2+3t+4}$$

Answer 2

Use the chain rule: $(g(f(x)))' = g'(f(x))*f'(x)$.

Your second step is an incorrect use of log rules. You cannot split that up. Try using the chain rule, and let me know if you need more help?

Answer 3

You are using logarithm rules incorrectly when splitting them up in the second line.

We apply the chain rule here.

Let $u=6t^2+9t+12$

$\frac{\mathrm{d} u}{\mathrm{d} t}=12t+9$

$\frac{\mathrm{d} }{\mathrm{d} u}(\ln (u))=\frac{1}{u}=\frac{1}{6t^2+9t+12}$

So $\frac{\mathrm{d} }{\mathrm{d} t}(\ln (6t^2+9t+12))=\frac{12t+9}{6t^2+9t+12}=\frac{4t+3}{2t^2+3t+4}$